Complex numbers answered questions that for centuries had puzzled the greatest minds in science. They are used to solve many scientific problems in the real world. Then, multiply through by $$r$$. Find the absolute value of $$z=\sqrt{5}−i$$. Express the complex number $4i$ using polar coordinates. Find the product and the quotient of ${z}_{1}=2\sqrt{3}\left(\cos \left(150^\circ \right)+i\sin \left(150^\circ \right)\right)$ and ${z}_{2}=2\left(\cos \left(30^\circ \right)+i\sin \left(30^\circ \right)\right)$. What does the absolute value of a complex number represent? Since this number has positive real and imaginary parts, it is in quadrant I, so the angle is . Converting complex number raised to a power to polar form. See Figure $$\PageIndex{7}$$. And then we have says Off N, which is two, and theatre, which is 120 degrees. The imaginary unit, denoted i, is the solution to the equation i 2 = –1.. A complex number can be represented in the form a + bi, where a and b are real numbers and i denotes the imaginary unit. √b = √ab is valid only when atleast one of a and b is non negative. See Example $$\PageIndex{6}$$ and Example $$\PageIndex{7}$$. Find the product and the quotient of $$z_1=2\sqrt{3}(\cos(150°)+i \sin(150°))$$ and $$z_2=2(\cos(30°)+i \sin(30°))$$. If $z=r\left(\cos \theta +i\sin \theta \right)$ is a complex number, then. Multiplying Complex numbers in Polar form gives insight into how the angle of the Complex number changes in an explicit way. To write complex numbers in polar form, we use the formulas $x=r\cos \theta ,y=r\sin \theta$, and $r=\sqrt{{x}^{2}+{y}^{2}}$. If ${z}_{1}={r}_{1}\left(\cos {\theta }_{1}+i\sin {\theta }_{1}\right)$ and ${z}_{2}={r}_{2}\left(\cos {\theta }_{2}+i\sin {\theta }_{2}\right)$, then the product of these numbers is given as: Notice that the product calls for multiplying the moduli and adding the angles. The absolute value $z$ is 5. For the following exercises, find all answers rounded to the nearest hundredth. $z=7\text{cis}\left(\frac{\pi}{6}\right)$, 18. Find the absolute value of the complex number $z=12 - 5i$. Example $$\PageIndex{2}$$: Finding the Absolute Value of a Complex Number with a Radical. Complex numbers can be added, subtracted, or … To convert from polar form to rectangular form, first evaluate the trigonometric functions. 45. From the origin, move two units in the positive horizontal direction and three units in the negative vertical direction. For the following exercises, write the complex number in polar form. Replace $$r$$ with $$\dfrac{r_1}{r_2}$$, and replace $$\theta$$ with $$\theta_1−\theta_2$$. 7.5 ­ Complex Numbers in Polar Form.notebook 1 March 01, 2017 Powers of Complex Numbers in Polar Form: We can use a formula to find powers of complex numbers if the complex numbers are expressed in polar form. Example $$\PageIndex{6B}$$: Finding the Rectangular Form of a Complex Number. The polar form of a complex number z = a + b ı is this: z = r(cos(θ) + ısin(θ)), where r = | z| and θ is the argument of z. Polar form is sometimes called trigonometric form as well. The rules are based on multiplying the moduli and adding the arguments. There are two basic forms of complex number notation: polar and rectangular. For the following exercises, find $\frac{z_{1}}{z_{2}}$ in polar form. 39. 37. If $\tan \theta =\frac{5}{12}$, and $\tan \theta =\frac{y}{x}$, we first determine $r=\sqrt{{x}^{2}+{y}^{2}}=\sqrt{{12}^{2}+{5}^{2}}=13\text{. Please support my work on Patreon: https://www.patreon.com/engineer4freeThis tutorial goes over how to write a complex number in polar form. 23. What is Complex Number? “God made the integers; all else is the work of man.” This rather famous quote by nineteenth-century German mathematician Leopold Kronecker sets the stage for this section on the polar form of a complex number. Use De Moivre’s Theorem to evaluate the expression. Plotting a complex number [latex]a+bi$ is similar to plotting a real number, except that the horizontal axis represents the real part of the number, $a$, and the vertical axis represents the imaginary part of the number, $bi$. 4 (De Moivre's) For any integer we have Example 4. In polar coordinates, the complex number $z=0+4i$ can be written as $z=4\left(\cos \left(\frac{\pi }{2}\right)+i\sin \left(\frac{\pi }{2}\right)\right)$ or $4\text{cis}\left(\frac{\pi }{2}\right)$. Express $$z=3i$$ as $$r\space cis \theta$$ in polar form. It states that, for a positive integer $$n$$, $$z^n$$ is found by raising the modulus to the $$n^{th}$$ power and multiplying the argument by $$n$$. Use the polar to rectangular feature on the graphing calculator to change $4\text{cis}\left(120^{\circ}\right)$ to rectangular form. Find the quotient of ${z}_{1}=2\left(\cos \left(213^\circ \right)+i\sin \left(213^\circ \right)\right)$ and ${z}_{2}=4\left(\cos \left(33^\circ \right)+i\sin \left(33^\circ \right)\right)$. 59. Evaluate the expression ${\left(1+i\right)}^{5}$ using De Moivre’s Theorem. 40. We use the term modulus to represent the absolute value of a complex number, or the distance from the origin to the point $$(x,y)$$. For the following exercises, find the absolute value of the given complex number. The first step toward working with a complex number in polar form is to find the absolute value. In order to work with complex numbers without drawing vectors, we first need some kind of standard mathematical notation. On the complex plane, the number $$z=4i$$ is the same as $$z=0+4i$$. We apply it to our situation to get. Notice that the moduli are divided, and the angles are subtracted. Find roots of complex numbers in polar form. $−\frac{1}{2}−\frac{1}{2}i$. For the following exercises, convert the complex number from polar to rectangular form. Find powers of complex numbers in polar form. $|z|=\sqrt{{x}^{2}+{y}^{2}}$, $\begin{array}{l}|z|=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ |z|=\sqrt{{\left(\sqrt{5}\right)}^{2}+{\left(-1\right)}^{2}}\hfill \\ |z|=\sqrt{5+1}\hfill \\ |z|=\sqrt{6}\hfill \end{array}$, $\begin{array}{l}|z|=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ |z|=\sqrt{{\left(3\right)}^{2}+{\left(-4\right)}^{2}}\hfill \\ |z|=\sqrt{9+16}\hfill \\ \begin{array}{l}|z|=\sqrt{25}\\ |z|=5\end{array}\hfill \end{array}$, $\begin{array}{l}x=r\cos \theta \hfill \\ y=r\sin \theta \hfill \\ r=\sqrt{{x}^{2}+{y}^{2}}\hfill \end{array}$, $\begin{array}{l}z=x+yi\hfill \\ z=r\cos \theta +\left(r\sin \theta \right)i\hfill \\ z=r\left(\cos \theta +i\sin \theta \right)\hfill \end{array}$, $\begin{array}{l}\hfill \\ x=r\cos \theta \hfill \\ y=r\sin \theta \hfill \\ r=\sqrt{{x}^{2}+{y}^{2}}\hfill \end{array}$, $\begin{array}{l}z=x+yi\hfill \\ z=\left(r\cos \theta \right)+i\left(r\sin \theta \right)\hfill \\ z=r\left(\cos \theta +i\sin \theta \right)\hfill \end{array}$, $\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ r=\sqrt{{0}^{2}+{4}^{2}}\hfill \\ r=\sqrt{16}\hfill \\ r=4\hfill \end{array}$, $\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ r=\sqrt{{\left(-4\right)}^{2}+\left({4}^{2}\right)}\hfill \\ r=\sqrt{32}\hfill \\ r=4\sqrt{2}\hfill \end{array}$, $\begin{array}{l}\cos \theta =\frac{x}{r}\hfill \\ \cos \theta =\frac{-4}{4\sqrt{2}}\hfill \\ \cos \theta =-\frac{1}{\sqrt{2}}\hfill \\ \theta ={\cos }^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{3\pi }{4}\hfill \end{array}$, $z=12\left(\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right)$, $\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}\\\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}$, $z=12\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)$, $\begin{array}{l}z=12\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)\hfill \\ \text{ }=\left(12\right)\frac{\sqrt{3}}{2}+\left(12\right)\frac{1}{2}i\hfill \\ \text{ }=6\sqrt{3}+6i\hfill \end{array}$, $\begin{array}{l}z=13\left(\cos \theta +i\sin \theta \right)\hfill \\ =13\left(\frac{12}{13}+\frac{5}{13}i\right)\hfill \\ =12+5i\hfill \end{array}$, $z=4\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)$, $\begin{array}{l}\hfill \\ \begin{array}{l}{z}_{1}{z}_{2}={r}_{1}{r}_{2}\left[\cos \left({\theta }_{1}+{\theta }_{2}\right)+i\sin \left({\theta }_{1}+{\theta }_{2}\right)\right]\hfill \\ {z}_{1}{z}_{2}={r}_{1}{r}_{2}\text{cis}\left({\theta }_{1}+{\theta }_{2}\right)\hfill \end{array}\hfill \end{array}$, $\begin{array}{l}{z}_{1}{z}_{2}=4\cdot 2\left[\cos \left(80^\circ +145^\circ \right)+i\sin \left(80^\circ +145^\circ \right)\right]\hfill \\ {z}_{1}{z}_{2}=8\left[\cos \left(225^\circ \right)+i\sin \left(225^\circ \right)\right]\hfill \\ {z}_{1}{z}_{2}=8\left[\cos \left(\frac{5\pi }{4}\right)+i\sin \left(\frac{5\pi }{4}\right)\right]\hfill \\ {z}_{1}{z}_{2}=8\left[-\frac{\sqrt{2}}{2}+i\left(-\frac{\sqrt{2}}{2}\right)\right]\hfill \\ {z}_{1}{z}_{2}=-4\sqrt{2}-4i\sqrt{2}\hfill \end{array}$, $\begin{array}{l}\frac{{z}_{1}}{{z}_{2}}=\frac{{r}_{1}}{{r}_{2}}\left[\cos \left({\theta }_{1}-{\theta }_{2}\right)+i\sin \left({\theta }_{1}-{\theta }_{2}\right)\right],{z}_{2}\ne 0\\ \frac{{z}_{1}}{{z}_{2}}=\frac{{r}_{1}}{{r}_{2}}\text{cis}\left({\theta }_{1}-{\theta }_{2}\right),{z}_{2}\ne 0\end{array}$, $\begin{array}{l}\frac{{z}_{1}}{{z}_{2}}=\frac{2}{4}\left[\cos \left(213^\circ -33^\circ \right)+i\sin \left(213^\circ -33^\circ \right)\right]\hfill \\ \frac{{z}_{1}}{{z}_{2}}=\frac{1}{2}\left[\cos \left(180^\circ \right)+i\sin \left(180^\circ \right)\right]\hfill \\ \frac{{z}_{1}}{{z}_{2}}=\frac{1}{2}\left[-1+0i\right]\hfill \\ \frac{{z}_{1}}{{z}_{2}}=-\frac{1}{2}+0i\hfill \\ \frac{{z}_{1}}{{z}_{2}}=-\frac{1}{2}\hfill \end{array}$, $\begin{array}{l}{z}^{n}={r}^{n}\left[\cos \left(n\theta \right)+i\sin \left(n\theta \right)\right]\\ {z}^{n}={r}^{n}\text{cis}\left(n\theta \right)\end{array}$, $\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ r=\sqrt{{\left(1\right)}^{2}+{\left(1\right)}^{2}}\hfill \\ r=\sqrt{2}\hfill \end{array}$, $\begin{array}{l}\tan \theta =\frac{1}{1}\hfill \\ \tan \theta =1\hfill \\ \theta =\frac{\pi }{4}\hfill \end{array}$, $\begin{array}{l}{\left(a+bi\right)}^{n}={r}^{n}\left[\cos \left(n\theta \right)+i\sin \left(n\theta \right)\right]\hfill \\ {\left(1+i\right)}^{5}={\left(\sqrt{2}\right)}^{5}\left[\cos \left(5\cdot \frac{\pi }{4}\right)+i\sin \left(5\cdot \frac{\pi }{4}\right)\right]\hfill \\ {\left(1+i\right)}^{5}=4\sqrt{2}\left[\cos \left(\frac{5\pi }{4}\right)+i\sin \left(\frac{5\pi }{4}\right)\right]\hfill \\ {\left(1+i\right)}^{5}=4\sqrt{2}\left[-\frac{\sqrt{2}}{2}+i\left(-\frac{\sqrt{2}}{2}\right)\right]\hfill \\ {\left(1+i\right)}^{5}=-4 - 4i\hfill \end{array}$, ${z}^{\frac{1}{n}}={r}^{\frac{1}{n}}\left[\cos \left(\frac{\theta }{n}+\frac{2k\pi }{n}\right)+i\sin \left(\frac{\theta }{n}+\frac{2k\pi }{n}\right)\right]$, $\begin{array}{l}{z}^{\frac{1}{3}}={8}^{\frac{1}{3}}\left[\cos \left(\frac{\frac{2\pi }{3}}{3}+\frac{2k\pi }{3}\right)+i\sin \left(\frac{\frac{2\pi }{3}}{3}+\frac{2k\pi }{3}\right)\right]\hfill \\ {z}^{\frac{1}{3}}=2\left[\cos \left(\frac{2\pi }{9}+\frac{2k\pi }{3}\right)+i\sin \left(\frac{2\pi }{9}+\frac{2k\pi }{3}\right)\right]\hfill \end{array}$, ${z}^{\frac{1}{3}}=2\left(\cos \left(\frac{2\pi }{9}\right)+i\sin \left(\frac{2\pi }{9}\right)\right)$, $\begin{array}{l}{z}^{\frac{1}{3}}=2\left[\cos \left(\frac{2\pi }{9}+\frac{6\pi }{9}\right)+i\sin \left(\frac{2\pi }{9}+\frac{6\pi }{9}\right)\right]\begin{array}{cccc}& & & \end{array}\text{ Add }\frac{2\left(1\right)\pi }{3}\text{ to each angle. Since De Moivre’s Theorem applies to complex numbers written in polar form, we must first write [latex]\left(1+i\right)$ in polar form. $z_{1}=2\sqrt{3}\text{cis}\left(116^{\circ}\right)\text{; }\left(118^{\circ}\right)$, 24. The absolute value of a complex number is the same as its magnitude, or $$| z |$$. Find roots of complex numbers in polar form. Use the polar to rectangular feature on the graphing calculator to change $5\text{cis}\left(210^{\circ}\right)$ to rectangular form. $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$, [ "article:topic", "DeMoivre\'s Theorem", "complex plane", "complex number", "license:ccby", "showtoc:no", "authorname:openstaxjabramson" ], https://math.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fmath.libretexts.org%2FBookshelves%2FAlgebra%2FBook%253A_Algebra_and_Trigonometry_(OpenStax)%2F10%253A_Further_Applications_of_Trigonometry%2F10.05%253A_Polar_Form_of_Complex_Numbers, $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$, Principal Lecturer (School of Mathematical and Statistical Sciences), 10.4E: Polar Coordinates - Graphs (Exercises), 10.5E: Polar Form of Complex Numbers (Exercises), Plotting Complex Numbers in the Complex Plane, Finding the Absolute Value of a Complex Number, Converting a Complex Number from Polar to Rectangular Form, Finding Products of Complex Numbers in Polar Form, Finding Quotients of Complex Numbers in Polar Form, Finding Powers of Complex Numbers in Polar Form, Finding Roots of Complex Numbers in Polar Form, https://openstax.org/details/books/precalculus. , but using a rational exponent r\space cis \theta\ ) in the real axis and the vertical axis the... 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